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# How to solve x^n = k?

x^n = k

I'm supposed to find all complex solutions. I know that for x^n = 1, it's just the nth roots of unity, but how would you solve it if the RHS wasn't 1

### 1 Answer

- Anonymous10 years agoFavorite Answer
It depends on what k can be. Is k always a positive integer? Is k always a negative integer? Can k be a complex number? We will need to consider all of these cases separately.

Case (i): k is a positive integer.

Since 1 = cos 0 + i sin 0, we see that, in polar form:

k = k(cos 0 + i sin 0).

By DeMovire's Theorem:

x = k^(1/n)

= [k(cos 0 + i sin 0)]^(1/n)

= k^(1/n) * [cos(2πk/n) + i sin(2πk/n)], where k = 0, 1, 2, ..., k-1.

Case (b): k is a negative integer.

Since -1 = cos π + i sin π, we see that:

k = |k|(cos π + i sin π).

Then, by DeMovire's Theorem:

x = k^(1/n)

= [|k|(cos π + i sin π)]^(1/n)

= |k|^(1/n) * {cos[(π + 2πk)/n] + i sin[(π + 2πk)/n]}, where k = 0, 1, 2, ..., k-1.

The last case can be done by just using DeMovire's Theorem on the complex number.

I hope this helps!